Problem: Simplify and expand the following expression: $ \dfrac{1}{3q + 9}+ \dfrac{2}{q + 3}+ \dfrac{q}{q^2 + 6q + 9} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{1}{3q + 9} = \dfrac{1}{3(q + 3)}$ We can factor the quadratic in the third term: $ \dfrac{q}{q^2 + 6q + 9} = \dfrac{q}{(q + 3)(q + 3)}$ Now we have: $ \dfrac{1}{3(q + 3)}+ \dfrac{2}{q + 3}+ \dfrac{q}{(q + 3)(q + 3)} $ The least common multiple of the denominators is: $ 3(q + 3)(q + 3)$ In order to get the first term over $3(q + 3)(q + 3)$ , multiply by $\dfrac{q + 3}{q + 3}$ $ \dfrac{1}{3(q + 3)} \times \dfrac{q + 3}{q + 3} = \dfrac{q + 3}{3(q + 3)(q + 3)} $ In order to get the second term over $3(q + 3)(q + 3)$ , multiply by $\dfrac{3(q + 3)}{3(q + 3)}$ $ \dfrac{2}{q + 3} \times \dfrac{3(q + 3)}{3(q + 3)} = \dfrac{6(q + 3)}{3(q + 3)(q + 3)} $ In order to get the third term over $3(q + 3)(q + 3)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{q}{(q + 3)(q + 3)} \times \dfrac{3}{3} = \dfrac{3q}{3(q + 3)(q + 3)} $ Now we have: $ \dfrac{q + 3}{3(q + 3)(q + 3)} + \dfrac{6(q + 3)}{3(q + 3)(q + 3)} + \dfrac{3q}{3(q + 3)(q + 3)} $ $ = \dfrac{ q + 3 + 6(q + 3) + 3q} {3(q + 3)(q + 3)} $ Expand: $ = \dfrac{q + 3 + 6q + 18 + 3q}{3q^2 + 18q + 27} $ $ = \dfrac{10q + 21}{3q^2 + 18q + 27}$